如何在Bash中将变量设置为命令的输出? [英] How do I set a variable to the output of a command in Bash?

问题描述

我有一个非常简单的脚本，类似于以下内容:

I have a pretty simple script that is something like the following:

#!/bin/bash

VAR1="$1"
MOREF='sudo run command against $VAR1 | grep name | cut -c7-'

echo $MOREF

当我从命令行运行此脚本并将参数传递给它时，我没有得到任何输出.但是，当我运行$MOREF变量中包含的命令时，我可以获取输出.

When I run this script from the command line and pass it the arguments, I am not getting any output. However, when I run the commands contained within the $MOREF variable, I am able to get output.

如何获取需要在脚本中运行的命令的结果，将其保存到变量中，然后在屏幕上输出该变量?

How can one take the results of a command that needs to be run within a script, save it to a variable, and then output that variable on the screen?

推荐答案

除了反引号`command`，来完成Command-Substitution"rel =" noreferrer>命令替换，我发现它们更易于阅读，并且可以嵌套.

In addition to backticks `command`, command substitution can be done with $(command) or "$(command)", which I find easier to read, and allows for nesting.

OUTPUT=$(ls -1)
echo "${OUTPUT}"

MULTILINE=$(ls \
   -1)
echo "${MULTILINE}"

引用(")对于保留多行变量值很重要；它在任务的右侧是可选的，例如

Quoting (") does matter to preserve multi-line variable values; it is optional on the right-hand side of an assignment, as word splitting is not performed, so OUTPUT=$(ls -1) would work fine.

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